Uniform continuity on a dense subset implies uniform continuity on the set.

Question

Let $f: (M_1,d_1)\to (M_2,d_2)$ be continous. If $f$ is uniformly continous on a dense subset $V$ of $M_1$, show $f$ is uniformly continous on $M_1$.

Could someone provide some hints? (no full answers).

I don't see how to start. How can I get from $\forall x,y\in M_1 = \overline V$ to $\forall x,y \in M_1$?

Answer 1

Use the contrapositive. So some $r > 0$ exists such that for all $\delta > 0$ we have $x(\delta), y(\delta) \in M_1$ such that $d(x(\delta), y(\delta)) < \delta$ and $d(f(x(\delta)), f(y(\delta))) \ge r$.

Now find $\delta_1$ for $s = \frac{r}{3} > 0$ on $V_1$ and see what you can do with continuity of $f$ at $x(\delta_1)$ and $y(\delta_1)$, and denseness of $V_1$.

Answer 2

You need to add the condition $(M_2, d_2)$ complet. For all element of $M_1-V$ you can use a Cauchy sequence converging to this element. The image by f of this Cauchy sequence is Cauchy in $(M_2, d_2)$, etc (here you need the completeness required). This answer has misunderstood the statement of the question.